# Thread: Series dealing with exponationals

1. ## Series dealing with exponationals

I don't see how

1. $\sum_{n=1}^\infty \frac{1}{e^n} + \frac{1}{n(n+1)}$ = $\frac{e}{e-1}$

2. $\sum_{n=1}^\infty e^\frac{1}{n} - e^\frac{1}{n+1}$ = $e-1$

In the first case I know that as $n\rightarrow\infty$ that $\frac{1}{e^n}$ approaches 0 and therefore it converges, and I know that $\frac{1}{n(n+1)} = 1$ (from a previous problem). But how do you know what $\frac{1}{e^n}$ sums up to?.

And in the second case I have no clue about approaching the problem, but I would think that it's the same type of approach as the first.

2. $
\sum\limits_{n = 1}^\infty {\left( {\frac{1}
{e}} \right)^n }$

is just an infinite geometric series, and we know that it converges since 1/e < 1 thus:

$\sum\limits_{n = 1}^\infty {\left( {\frac{1}
{e}} \right)^n } = \frac{{a_1 }}
{{1 - q}} = \frac{{\frac{1}
{e}}}
{{1 - \frac{1}
{e}}} = \frac{1}
{{e - 1}}$

the second term can be rewritten using partial fraction decomposition as:

$
\sum\limits_{n = 1}^\infty {\frac{1}
{{n\left( {n + 1} \right)}} = } \sum\limits_{n = 1}^\infty {\frac{1}
{n}} - \frac{1}
{{n + 1}}$

$
\sum\limits_{n = 1}^\infty {\frac{1}
{n}} - \frac{1}
{{n + 1}} = \left( {1 - \frac{1}
{2}} \right) + \left( {\frac{1}
{2} - \frac{1}
{3}} \right) + \left( {\frac{1}
{3} - \frac{1}
{4}} \right) + \cdots = 1
$

since both individual terms converge to 0.
so:

$
\sum\limits_{n = 1}^\infty {\frac{1}
{{e^n }}} + \sum\limits_{n = 1}^\infty {\frac{1}
{{n\left( {n + 1} \right)}} = \frac{1}
{{e - 1}} + 1 = \frac{e}
{{e - 1}}}
$

$
\sum\limits_{n = 1}^\infty {e^{\frac{1}
{n}} - e^{\frac{1}
{{n + 1}}} = \left( {e - e^{\frac{1}
{2}} } \right) + \left( {e^{\frac{1}
{2}} - e^{\frac{1}
{3}} } \right) + \cdots }
$

now let's look at the sum of the first N elements and then take the limit as N approaches infinity:

$
\mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {e^{\frac{1}
{n}} - e^{\frac{1}
{{n + 1}}} = \left( {e - e^{\frac{1}
{2}} } \right) + \left( {e^{\frac{1}
{2}} - e^{\frac{1}
{3}} } \right) + \cdots }
$
$
+ \left( {e^{\frac{1}
{{N - 1}}} - e^{\frac{1}
{N}} } \right) + \left( {e^{\frac{1}
{N}} - e^{\frac{1}
{{N + 1}}} } \right) = e - 1
$